## anonymous one year ago Simplify the rational expression. State any restrictions on the variable.

1. anonymous

$\frac{ n^4-10n^2+24 }{ n^4-9^2+18 }$

2. math&ing001

First lets find the restrictions. For that we solve for n : n^4 - 9^2 + 18 = 0

3. anonymous

Than what? @math&ing001

4. anonymous

5. anonymous

6. anonymous

7. anonymous

8. anonymous

@Hero

9. anonymous

@satellite73

10. Hero

Let $$n^2 = y$$ then try simplifying the expression.

11. anonymous

Please help me I know its $\frac{ n^2-4 }{ n^2-3}$ but i dont know the other half for my answer, so it's between A-C for the answer

12. anonymous

(n+2)(n−2)(n2−6) (n2−3)(n2−6)

13. Hero

How do you know it's $$\dfrac{n^2 - 4}{n^2 - 3}$$?

14. anonymous

(n+2)(n−2)(n2−6) (n2−3)(n2−6)

15. Hero

Multiply it back out to see if what you get matches the original expression.

16. Hero

Also the original expression you posted likely has a typo in it somewhere.

17. anonymous

How do i do that?

18. anonymous

@timo86m

19. anonymous

@Astrophysics

20. anonymous

21. anonymous

@mathmate

22. anonymous

I just need help with the end of the problem. I think the answer is B though

23. mathmate

@EllenJaz17 Please confirm that the problem is indeed: $$\Large \frac{ n^4-10n^2+24 }{ n^4-9\color{red}{n}^2+18 }$$

24. anonymous

yes

25. anonymous

Yes it is @mathmate

26. Michele_Laino

hint: we can factorize the denominator, using the same procedure for numerator, so we can write this: ${n^4} - 9{n^2} + 18 = \left( {{n^2} - 6} \right)\left( {{n^2} - 3} \right)$

27. anonymous

(n+2)(n−2)(n2−6) (n2−3)(n2−6)

28. anonymous

$\frac{ n^2-4 }{ n^2-3 }$

29. Michele_Laino

yes!

30. anonymous

I know thats the answer to the first half but i need to find out the second half

31. anonymous

32. anonymous

33. anonymous

34. anonymous

35. Michele_Laino

the denominator is equal to zero when these two conditions hold: $\begin{gathered} {n^2} - 6 = 0 \hfill \\ \hfill \\ {n^2} - 3 = 0 \hfill \\ \end{gathered}$

36. Michele_Laino

please remember, that we can not divide by zero

37. anonymous

correct

38. Michele_Laino

39. anonymous

6 and 3?

40. Michele_Laino

are you sure? we have 6^2= 36, and 3^2=9

41. anonymous

No im not sure im just guessing

42. Michele_Laino

hint: the solution to this equation: ${n^2} - k = 0$ are: $n = \sqrt k ,\quad n = - \sqrt k$

43. Michele_Laino

solutions*

44. Michele_Laino

since: ${\left( {\sqrt k } \right)^2} = {\left( { - \sqrt k } \right)^2} = {n^2}$

45. Michele_Laino

being k a positive number or k=0

46. anonymous

47. Michele_Laino

no, I don't think so, sorry!

48. anonymous

49. Michele_Laino

hint: what are the solution of this equation: ${n^2} - 3 = 0$ ?

50. anonymous

So that only leaves A & C correct? because D isn't $\frac{ n^2-4 }{ n^2-3 }$

51. Michele_Laino

solutions*

52. anonymous

$n^{2}=3?$

53. Michele_Laino

yes! and what is n=...?

54. anonymous

$n \neq \pm \sqrt{6}\pm \sqrt{3}?$

55. Michele_Laino

from this equation: ${n^2} - 3 = 0$ I get $n = \pm \sqrt 3$ am I right?

56. anonymous

yes

57. Michele_Laino

ok! so we have to exclude those 2 values, since when: $n = \pm \sqrt 3$ the denominator is zero Now do the same with the equation: ${n^2} - 6 = 0$

58. anonymous

$n=\pm \sqrt{6}$

59. Michele_Laino

perfect, we have to exclude those values too, since when: $n = \pm \sqrt 6$ the denominator is zero. So what is the right option?

60. anonymous

A

61. Michele_Laino

that's right! :)

62. anonymous

Thank you so much for teaching me how to get to the answer!

63. Michele_Laino

:)

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